Let $R$ be a commutative ring with ideal $\mathfrak m$ satisfying $\mathfrak m^2 = \mathfrak m$, and let $f:M\to N$ be a map of $R$-modules. I want to show that the kernel and cokernel of $f$ are annihilated by $\mathfrak m$ if and only if we get an isomorphism after tensoring with $\widetilde{\mathfrak m} = \mathfrak m\otimes_R \mathfrak m$. Any pointers appreciated! Thanks!
$\textbf{Edit}$: the implications on cokernels in both directions are straightforward using the right exactness of the tensor product; it's the kernels I'm having a bit of trouble with. For example, given that $f\otimes \widetilde{\mathfrak{m}}$ is an isomorphism, to show that $(\ker f)\otimes \mathfrak m = 0$, I would like to compare the Tor exact sequence:$$…\to \widetilde{\mathfrak{m}}\otimes \ker f\to \widetilde{\mathfrak{m}}\otimes M\xrightarrow{\widetilde{\mathfrak{m}}\otimes f}\widetilde{\mathfrak{m}}\otimes \operatorname{im}(f)\to 0 $$ with the short exact sequence $$0 \to \ker(\widetilde{\mathfrak{m}}\otimes f)\to \widetilde{\mathfrak{m}}\otimes M\xrightarrow{\widetilde{\mathfrak{m}}\otimes f} \widetilde{\mathfrak{m}}\otimes N\to 0$$ The issue is that I don't know if the map from $\operatorname{Tor}^R_1(\widetilde{\mathfrak{m}}, \operatorname{im}(f))$ vanishes…similarly, for the converse, given that $\ker f $ and $\operatorname{coker} f$ are annihilated by $\mathfrak{m}$, to get $\ker(\widetilde{\mathfrak{m}}\otimes f) = 0$, I would like to compare the same two exact sequences; now the problem seems to be $\operatorname{Tor}^R_1(\widetilde{\mathfrak{m}} , \operatorname{coker}(f))$. Perhaps I'm going about this all wrong. Any pointers much appreciated!
Best Answer
Define $\widetilde{f}=1_{\widetilde{\mathfrak{m}}}\otimes f.$
The converse direction is quite easy, and works just as well for $\mathfrak m$ as $\widetilde{\mathfrak{m}}.$
Assume $\widetilde{f}$ is an isomorphism. Consider an arbitrary $x\in\ker f$ and $m\in\mathfrak{m}.$ We want to show that $mx=0.$ (For the cokernel we can just use right exactness, as you mention.) Write $m$ as a finite sum $\sum a_ib_i$ with $a_i,b_i\in\mathfrak{m}.$ Define $x'=\sum a_i\otimes b_i\otimes x\in \widetilde{\mathfrak{m}}\otimes M.$ Then $\widetilde{f}(x')=\sum a_i\otimes b_i\otimes f(x)=0$ because $x\in\ker f.$ Since $\widetilde{f}$ is an isomorphism, $x'$ is zero. Let $\mu$ be the multiplication map $\widetilde{\mathfrak{m}}\otimes M\to M$ (defined on simple tensors by $\mu(a\otimes b\otimes y)=aby$). Then $mx=\mu(x')=\mu(0)=0.$
For the forward direction, assume that $\mathfrak{m}$ annihilates $\ker f$ and $\operatorname{coker} f.$
First step: reduce to the case that $f$ is injective. To do this, factorize $f$ as $M\to M/\ker f \to N.$ Observe that $\mathfrak{m}\otimes \ker f$ is zero: it's generated by simple tensors $a\otimes x$ with $a\in\mathfrak m$ and $x\in\ker f,$ and by picking a decomposition $a=\sum a_ib_i\otimes x$ with $a_i,b_i\in\mathfrak m$ we get $a\otimes x=\sum a_i\otimes b_ix=0.$ So $\widetilde{\mathfrak{m}}\otimes \ker f$ is also zero. Since $\widetilde{\mathfrak{m}}\otimes -$ is right exact, it sends $M\to M/\ker f$ to an isomorphism. So we just need to prove that it sends the image map $M/\ker f\to N$ to an isomorphism. From here on, let $f$ denote this image map. Note this doesn't change the cokernel.
Let's try to define an inverse $$g:\widetilde{\mathfrak{m}}\otimes N\to \widetilde{\mathfrak{m}}\otimes M$$ to $\widetilde{f}.$ We will do this by defining a multilinear function $G:\mathfrak m\times\mathfrak m\times N \to \widetilde{\mathfrak{m}}\otimes M,$
For $y\in N$ and $m\in\mathfrak m$ let $f^{-1}(my)$ denote the unique element $x\in M$ such that $f(x)=my,$ which exists by the assumption that $m$ annihilates $\operatorname{coker} f.$ To define $G(a,b,y)$ pick a decomposition $a=\sum_i a_ia'_i$ with $a_i,a'_i\in\mathfrak{m},$ and set $G(a,b,y)=\sum_i (a_i\otimes b\otimes f^{-1}(a'_iy)).$
To see that $G$ doesn't depend on the choice of decomposition of $a,$ consider a decomposition $b=\sum b_jb'_j$ with $b_j,b'_j\in\mathfrak m.$ The condition on $x$ is equivalent to \begin{align*} x&=\sum_{i,j} (a_i\otimes b_j\otimes b'_jf^{-1}(a'_ix))\\ &=\sum_{i,j} (a_i\otimes b_j\otimes a'_if^{-1}(b'_jx))\\ &=\sum_j (a\otimes b_j\otimes f^{-1}(b'_jx)) \end{align*} which is independent of the decomposition of $a.$
Define $g$ by $g(a\otimes b\otimes y)=G(a,b,y).$ There are a few things to check but I think they are all easy from here: